Binomial theorem to evaluate limits?

In summary: I do not remember the details of the long division process. I think I just factored out "x" in the numerator and cancelled it with the denominator. I do remember just seeing "x" in the divisor and "x" in the numerator and thinking "1". I do remember that can be a good way to get to the end quickly. But when it does not work, then you have to do something else. The other way I remember is that if you miss a term from the numerator, that is to say if you have "1" in the divisor, then you can miss "1" from the numerator. I think that was the case here. In that case, you are just
  • #1

Homework Statement



lim x->1 (X^9 + x -2)/(x^4 + x -2)

I know how to do this using L'Hopitals Rule and I get 2

Homework Equations



(1+b)^n = 1 + bn + n(n-1)b^2/2! + n(n-1)(n-2)b^3/3! ...

The Attempt at a Solution



Let x = h+1

x -> 1
h -> 0

lim h->0 (h+1)^9 + h-1/(h+1)^4+h-1

lim h->0 h^9 (1+1/h)^9 +h-1/h^4(1+1/h)^4 +h-1

Binomial theorem:

(1+1/h)^9 = 1 +9/h +36/h^2 + 84/h^3 ...

(1+1/h)^4 = 1 +4/h + 6/h^2 + 4/h^3 ...

lim h->0 h^9(1 +9/h +36/h^2 + 84/h^3 ...) +h-1/h^4(1 +4/h + 6/h^2 + 4/h^3 ...) +h-1


This is stil ∞.0 right?

I tried to get rid of the +h-1 at the end by doing

h^7(h^2 +10h +35 +84/h...)/h^2(h^2+5h+5+4/h...)

but then you get ∞.∞
 
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  • #2
You are fine to
lim h->0 ((h+1)^9 + h-1)/((h+1)^4+h-1)
But I don't understand why you then switched to "1/h". Taking h to 0 is much simpler than taking 1/h to infinity. By the binomial theorem, (h+ 1)^9= h^9+ 9h^8+ ...+ 36h^2+ 9h+ 1 so that (h+1)^9+ h- 1= h^9+ 9h^8+ ...+ 36h^2+ 10h. Similarly (h+ 1)^4= h^4+ 4h^3+ 6h^2+ 4h+ 1 so that (h+ 1)^4+ h- 1= h^4+ 4h^3+ 6h^2+ 5h.

The fraction can be written
[tex]\frac{h^9+ 9h^9+ ...+ 36h^2+ 10h}{h^4+ 4h^3+ 6h^2+5h}= \frac{h(h^8+ 9h^7+ ...+ 36h+ 10)}{h(h^3+ 4h^2+ 6h+ 5)}[/tex]
Now those first "h" terms cancel and you can just take h= 0.

Are you required to use that method? Simpler than that or L'Hopital:

The only problem here is that both numerator and denominator are 0 at x= 1. But since they are polynomials, that means that each has x- 1 as a factor:
[tex]x^9+ x- 2= (x- 1)(x^8+ x^7+ x^6+ x^5+ x^4+ x^3+ x^2+ x+ 2)[/tex]
and
[tex]x^4+ x- 2= (x- 1)(x^3+ x^2+ x+ 2)[/tex]
So you just need to evaluate
[tex]\frac{x^8+ x^7+ x^6+ x^5+ x^4+ x^3+ x^2+ x+ 2}{x^3+ x^2+ x+ 2}[/tex]
at x= 1.
 
  • #3
HallsofIvy said:
You are fine to
lim h->0 ((h+1)^9 + h-1)/((h+1)^4+h-1)
But I don't understand why you then switched to "1/h". Taking h to 0 is much simpler than taking 1/h to infinity. By the binomial theorem, (h+ 1)^9= h^9+ 9h^8+ ...+ 36h^2+ 9h+ 1 so that (h+1)^9+ h- 1= h^9+ 9h^8+ ...+ 36h^2+ 10h. Similarly (h+ 1)^4= h^4+ 4h^3+ 6h^2+ 4h+ 1 so that (h+ 1)^4+ h- 1= h^4+ 4h^3+ 6h^2+ 5h.

The fraction can be written
[tex]\frac{h^9+ 9h^9+ ...+ 36h^2+ 10h}{h^4+ 4h^3+ 6h^2+5h}= \frac{h(h^8+ 9h^7+ ...+ 36h+ 10)}{h(h^3+ 4h^2+ 6h+ 5)}[/tex]
Now those first "h" terms cancel and you can just take h= 0.

Are you required to use that method? Simpler than that or L'Hopital:

The only problem here is that both numerator and denominator are 0 at x= 1. But since they are polynomials, that means that each has x- 1 as a factor:
[tex]x^9+ x- 2= (x- 1)(x^8+ x^7+ x^6+ x^5+ x^4+ x^3+ x^2+ x+ 2)[/tex]
and
[tex]x^4+ x- 2= (x- 1)(x^3+ x^2+ x+ 2)[/tex]
So you just need to evaluate
[tex]\frac{x^8+ x^7+ x^6+ x^5+ x^4+ x^3+ x^2+ x+ 2}{x^3+ x^2+ x+ 2}[/tex]
at x= 1.

How did you divide x^4 +x -2 by (1-x)? When I did it I got (-x^3+x^2+x) -2/(1-x)
I did the polynomial division method.

Also, thank you.
 
  • #4
applestrudle said:
How did you divide x^4 +x -2 by (1-x)? When I did it I got (-x^3+x^2+x) -2/(1-x)
I did the polynomial division method.

Also, thank you.

Sorry, I mean how did you divide x^4 +x -2 by (x-1)? I got (x^3 -x^2 -x -2/(x-1) )

and for x^9 +x -2 divided by (x-1) I got (x^8-x^7 -x^6 -x^5 -x^4 -x^3 -x^2 -x -2/(x-1))
 
  • #5
applestrudle said:
Sorry, I mean how did you divide x^4 +x -2 by (x-1)? I got (x^3 -x^2 -x -2/(x-1) )

and for x^9 +x -2 divided by (x-1) I got (x^8-x^7 -x^6 -x^5 -x^4 -x^3 -x^2 -x -2/(x-1))

Whether they are true is something you can more easily check than how it was got.
 

1. What is the Binomial Theorem?

The Binomial Theorem is a mathematical formula that describes the expansion of powers of binomials. It states that the coefficient of the kth term in the expansion of (a + b)n is given by the formula: n!/(k!(n - k)!), where n is a positive integer and a and b are any real numbers.

2. How is the Binomial Theorem used to evaluate limits?

The Binomial Theorem can be used to evaluate limits by expanding a binomial expression and then taking the limit as the number of terms increases. This is particularly useful for limits involving radicals or fractional exponents.

3. What is the process for using the Binomial Theorem to evaluate limits?

The process for using the Binomial Theorem to evaluate limits involves first identifying the binomial expression in the limit and then expanding it using the formula mentioned in the first question. Next, the limit is taken as the number of terms in the expansion approaches infinity. Finally, any remaining algebraic simplifications can be made to determine the final limit value.

4. What are some common mistakes when using the Binomial Theorem to evaluate limits?

Some common mistakes when using the Binomial Theorem to evaluate limits include not correctly identifying the binomial expression, not expanding the expression correctly, and not taking the limit as the number of terms approaches infinity. It is important to carefully follow the steps and be mindful of algebraic errors.

5. Can the Binomial Theorem be used to evaluate limits for all types of functions?

No, the Binomial Theorem is specifically used for evaluating limits involving binomial expressions. It may not be applicable for other types of functions, such as trigonometric or exponential functions. In these cases, other techniques such as L'Hôpital's rule or algebraic manipulation may be more useful.

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